write program to perform sum = 1+ (1+2) + (1+2+3) + ... + (1+2...+n)

I can't get the codes right. Can somebody help?

#include<stdio.h>
int main()
{
 int n, sum,i,j;

 printf("Please enter an integer, n = ");
 scanf("%d", &n);

 for(i=1;i<=n;开发者_JAVA技巧i++)
     for(j=1;j<=i;j++)
         sum = sum + n;
 printf("sum = %d", sum);


 return 0;
}

1. You are not initialising sum. Initialise it with 0.
2. You shouldn't be adding n at each step, but j.

Of course, this is to fix your current code. There are better approaches to solving the problem, which others have already mentioned.

Edit:

Just for fun, here's a formula that allows you to solve the problem in O(1):

Your sum is equal to n*(n + 1)*(2*n + 1) / 12 + n*(n + 1) / 4.

This is obtained by writing it as a sum and using the fact that the sum of the first n consecutive squares is n(n + 1)(2n + 1) / 6 and the sum of the first n positive ints is n(n + 1)/2. +1 if you can find a nicer form of the formula.

No need for recursion, just look at the math:

1 + (1+2) + (1+2+3) + ... + (1+2+3+...+n)

is equal to

1*n + 2*(n-1) + 3*(n-2) + ... + n

Not what you expected, but this is the best solution ;)

int calculate (int n) {
  return (2*n + 3*n*n + n*n*n) / 6;
}

Think this through. You have one sum you're accumulating, and you have a series of values. Each value can be generated from the previous one by adding the index. So why do you have nested loops?

Doing it iteratively, like you tried:

#include <stdio.h>

int main() {
    int i, t, n, sum;
    printf("Please enter an integer, n = ");
    scanf("%d", &n);
    t = sum = 0;
    for (i = 1; i <= n; ++i) {
        t += i;
        sum += t;
    }
    printf("sum = %d\n", sum);
    return 0;
}

But there's a closed-form formula as IVlad suggested.

You never initialize sum, so you're adding everything to a random garbage value. Stick sum = 0; before your loops

Start with the math, see if you can find some pattern.

if n = 0 , res = 0?
if n = 1 , res = 1
if n = 2 , res = 1 + (1+2) 
if n = 3 , res = 1 + (1+2) + (1+2+3)

for each n, res is ??

Try:

int main(void)
{
    int     n, sum;

    printf("\nPlease enter a postive integer value");    
    scanf("%d", &n);
    sum = n * ( n + 1 )/ 2;
    printf("\n%d", sum);
    return 0;
}

Try this:

int main(void)
 {
   int total=1;
   int sumtotal = 0;
   int n=5;

   for(int i=1; i<=n; i++)
     {
       total+=i;
       sumtotal+=total;
     }
   //sumtotal should give you 1+(1+2)+(1+2+3)
   return 0;
 }

n * (n -(n - 1))

int n, sum 

for n = 0; n <= 3; n ++ {
    sum += n * (n -(n - 1))
 }

works for factorials too (change operator and set sum to 1):

int n, sum 

sum = 1

for n = 0; n <= 3: n++{
    sum *= n * (n -(n -1))
}